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#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.4 question 6

$\frac{d y}{d x}=\frac{y-x^{4}}{y^{4}-x}$

Hint:

Use chain rule to find the differentiation formula like $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$

Given:

$x^{5}+y^{5}=5 x y$

Solution:

Differentiate the given equation w.r.t x

$\frac{d}{d x}\left(x^{5}+y^{5}\right)=\frac{d}{d x}(5 x y)$

$\frac{d}{d x}\left(x^{5}\right)+\frac{d}{d x}\left(y^{5}\right)=5 \frac{d(x y)}{d x}$

$5 x^{4}+\left(\frac{d y^{5}}{d y} \times \frac{d y}{d x}\right)=5\left[x \frac{d y}{d x}+y \frac{d x}{d x}\right]$                    $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}, \frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$

$5 x^{4}+5 y^{4} \frac{d y}{d x}=5\left[x \frac{d y}{d x}+y\right]$

$5 x^{4}+5 y^{4} \frac{d y}{d x}=5 x \frac{d y}{d x}+5 y$

$5 y^{4} \frac{d y}{d x}-5 x \frac{d y}{d x}=5 y-5 x^{4}$

$\frac{d y}{d x}\left(5 y^{4}-5 x\right)=5 y-5 x^{4}$

$\frac{d y}{d x}=\frac{5 y-5 x^{4}}{5 y^{4}-5 x}$

\begin{aligned} &\frac{d y}{d x}=\frac{5\left(y-x^{4}\right)}{5\left(y^{4}-x\right)} \\ &\frac{d y}{d x}=\frac{y-x^{4}}{y^{4}-x} \end{aligned}

Hence  $\frac{d y}{d x}=\frac{y-x^{4}}{y^{4}-x}$  is the required answer